Question

a trough has ends shaped like isosceles triangles, and has a length of 9 m, width 3 m, and height 4 m, as shown in the image below. water is being pumped into the trough at a rate of 3 m³/min. at what rate does the height of the water change when the water is 2 m deep? note that the volume of a trough with length $l$, width $w$, and height $h$ is given by $v=\frac{1}{2}cdot lcdot wcdot h$. submit an exact answer.

Explanation:

Step1: Relate width and height of water

Since the cross - section is an isosceles triangle, by similar triangles, the ratio of width to height of the whole trough is $\frac{3}{4}$. For the water in the trough, if the height of water is $h$ and width is $w$, then $\frac{w}{h}=\frac{3}{4}$, so $w = \frac{3}{4}h$. The length of the trough $l = 9$. The volume of water $V=\frac{1}{2}lwh=\frac{1}{2}\times9\times\frac{3}{4}h\times h=\frac{27}{8}h^{2}$.

Step2: Differentiate volume with respect to time

Differentiate $V=\frac{27}{8}h^{2}$ with respect to time $t$ using the chain - rule. $\frac{dV}{dt}=\frac{27}{4}h\frac{dh}{dt}$.

Step3: Substitute known values

We know that $\frac{dV}{dt}=3$ (the rate of water being pumped in) and $h = 2$. Substitute these values into the equation $\frac{dV}{dt}=\frac{27}{4}h\frac{dh}{dt}$. So, $3=\frac{27}{4}\times2\times\frac{dh}{dt}$.

Step4: Solve for $\frac{dh}{dt}$

First, simplify the right - hand side of the equation: $\frac{27}{4}\times2\times\frac{dh}{dt}=\frac{27}{2}\frac{dh}{dt}$. Then solve for $\frac{dh}{dt}$: $\frac{dh}{dt}=\frac{3\times2}{27}=\frac{2}{9}$ m/min.

Answer:

$\frac{2}{9}$ m/min